Find the resistance (in Ω) that must be placed in parallel with a 25.0 Ω galvanometer having a 50.0 µA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 540 mA full-scale reading.

2.315×10⁻³ Ω

Explanation:

To convert a galvanometer to an ammeter, a low resistance (Shunt) is connected in parallel to the galvanometer.

From the question,

Voltage across the parallel resistor = voltage across the resistance of the galvanometer.

From ohm's law,

Voltage = Current×Resistance.

I'R' = IR Equation 1

Where I' = current flowing through the galvanometer, R' = Resistance of the galvanometer, I = current flowing through the the parallel resistor, R = Resistance of the parallel resistor.

make R the subject of the equation

R = I'R'/I Equation 2

Given: I' = 50 μA = 50×10⁻⁶ A, R' = 25 Ω, I = (540 mA -  50 μA) = 0.54 - 0.00005 = 0.53995 A.

Substitute into equation 2

R = 50×10⁻⁶(25)/0.53995

R = 2.315×10⁻³ Ω

Hence the the resistance in parallel = 2.315×10⁻³ Ω

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